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3 years ago

The frequency table shows the scores from rolling a dice.

Mathematics

2 answers:

Tatiana [17]3 years ago

5 0

The answer is 3 (or 3.08)

Crank3 years ago

5 0

U add 1+1+2+2+3+1+4+2+5+4+6+2 which is 33 and divide it by how many numbers is in the data set which is 12. Then do 33 divided by 12 which is 3

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In trapezoid EFGH, EF ≅ HG. What is the measure of the angle between Elm Street and Pine Avenue?

m_a_m_a [10]

Angle E is 72 degrees

6 0

3 years ago

Read 2 more answers

For 9-20 tell whether the first number is a multiple of the second number

jeyben [28]

Answer:

21,7 - yes

28,3 - no

17,3 - no

20,4 - yes

55,5 - yes

15,5 - yes

26,4 - no

32,8 - yes

48,7 no

60,2 - yes

79,4 - no

81,3 - yes

Step-by-step explanation:

hope that helps

7 0

2 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

3 years ago

A kitten's weight increase by 12% each week for 4 mouths the little currently weighs 2.6 pounds which will not calculate how muc

krok68 [10]

Answer:

a. 2.6(0.12)

Step-by-step explanation:

Check all of the other answers: b. c. and d. are variations of equations that equal to 2.912. 2.6(0.12) is not the correct equation because it equals to 0.312. The kitten‘s weight must increase, so that is the incorrect equation.

5 0

3 years ago

What are the values for the coefficients and constant term of 0 = 2 + 3x2 - 5x?

pav-90 [236]

Answer:

Step-by-step explanation:

a=3

b=-5

c=2

7 0

3 years ago